A very tricky trig question?

Question

Hari wants to set up a 1500m obstacle course.
The start-finish line is at grid reference 251132. The first leg is 600m long. First turn is at grid reference 631596. Second leg is 250m long at a bearing of 309 degrees.
Is this a 1500metre course?

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Answer 1

You are asking us to make too many assumptions. You have given no explanation for how to understand the grid reference system you are using. Without this information, we cannot tell the bearing of the first leg. You have also not specifically told us that the turns are arcs of circles or what their radii are.

Answer 2

631 - 251 = 380
596 - 132 = 464
atan(380/464) = 39.316°
309° - 180° - 39.316° = 89.684°
d3^2 = 360,000 + 62,500 - 1,656.1
d3^2 = 420,843.8987m^2
d3 = 648.725 m
Σ = 1,498.725 m
No, it is 1.275 m short

Answer 3

600 and 250 is the displacement given
the distance can be 1500


thank u thank u..

Answer 4

?(a million/2)?[a million+sin?]²d? = ?(a million/2)?[a million + 2sin? + sin²?]d? = ??/2 + ?sin? + (?/2)sin²? d? ??/2d? = ?²/4 ??sin?d? = -?cos? + ?cos?d? = -?cos? + sin? ?(?/2)sin²?d?: cos2? = cos²? - sin²? = a million - 2sin²? sin²? = (a million - cos2?)/2 ?(?/2)sin²?d? = ?(?/4)(a million - cos2?)d? = (?/4)(? - (a million/2)sin2?) - ?(a million/4)(? - (a million/2)sin2?)d? = (?/4)(? - (a million/2)sin2?) - (a million/4)(?²/2) + (a million/16)cos2?) = ?²/4 - (?/8)sin2? - ?²/8 - (a million/16)cos2? = ?²/8 - (?/8)sin2? - (a million/16)cos2? ??/2 + ?sin? + (?/2)sin²? d? = ?²/4 - ?cos? + sin? + ?²/8 - (?/8)sin2? - (a million/16)cos2? = 3?²/8 - ?cos? + sin? - (?/8)sin2? - (a million/16)cos2? prepare the limitations to get the superb answer

Answer 5

no